My question is: "Is it possible to have a sound and rigorous legitimation of the following construction ?". This construction is:

0/ Let ZFC be the usuel set theory, and let us add to the language capital latin letters as names for classes.Let V={x/x=x} be the usual universal class (that is a proper class, directly by foundation, or by Russel's paradox and separation) and P(x)={y/y⊆x} be the usual Power set fonction.

1/ Let us now extend the domain of the power set fonction to all classes, so that P(A)={y/y⊆A} We have that P(A)is a set if A is a set by the power set axiom;but as F(x)={x} in an injection from A into P(A), we also hav}e that P(A) must be a proper class if A is such. So that P(A) is a set iff A is a set. The universal class V being transitive and well-founded, we have that V={x/x⊆V}={x/x⊆V}=P(V); So that: V=P(0,V)=P(V)=P(1,V)=P(P(n,V))=P(n+1,V). Let us also remark that for every class A, we have A⊆V, so tat V is the greater possible class.

2/ Now, let us recursively build the following Q construction: Q(0,V)=P(V)=V{x/x⊆V}=V and Q(n+1,V)={A/(B∈A-->B∈Q(n,V)}={A/A Q(n,V)} Moreover let the collection Q be the union of the collections Q(n,V): Q=∪〈n∈N〉Q(n,V);

By induction, we have Q(n,V)⊆Q(n+1,V). We already know that the elements of Q(0,V)=V are axactly sets, and that the elements of Q(1,V)={A/A⊆V} are exactly usual classes. So that Q(0,V)⊆Q(1,V); moreover, the elements of the difference Q(1,V)/Q(0,V) are exactly usual proper classes and V is such an element. We also have Q(n,V)⊆Q(n+1,V)-->Q(n+1,V)⊆Q(n+2,V) because A∈Q(n+1,V)<-->(∀B∈A(B∈Q(n,V)) and by hypothesis B∈Q(n,V)-->B∈Q(n+1,V), so that A∈Q(n+2,V).

It is now natural to name sets as level 0 super-classes, proper classes as level 1 super-classes, and more generally elements of the difference Q(n+1,V)/Q(n,V) as level (n+1) super-classes, and also members of Q as super-classes.

We already know that Q(0,V) and Q(1,V)/Q(0,V) are non-void. more generally we have that Q(n+1,V)/Q(n,V) is non-void. To see this, let A and B be two level n super-classes and consider R(A,B)={D∈Q(n,V)/A⊆D⊆B}. It is clear that R(A,B) is a level (n+1) super-class that is non-void as soon as A⊆B. Particularly R(∅,A)={B/B⊆A} and R(A,A)={A} are such level (n+1) super-classes. So, we have by induction that Q(n,V)⊆Q(n+1,V) and also that Q(n,V)∈Q(n+1,V); for every level n, Q(n,V) is the greater element of Q(n+1,V) for inclusion.

3/ We also obtain that the difference of two super-classes (of level n) is a super-class (of level n), and that the union (the intersection) of a finite (of any family, but what can be the collection of indices ?) family of super-classes (of level n) is a super-class (of level n; of level ≤n).

Gérard Lang

`$`

symbols (and`$$`

for centered display math), but you need to sometimes add spaces, or use`\lbrace`

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, to avoid conflicts with the parser. – Asaf Karagila Sep 30 '12 at 16:08